Why does stress build up at transform boundaries?

1 answer:

3 0

Tectonic plate edges are not smooth. At transform boundaries, plates move against each other, but since the plates aren't smooth, lockups form. Stress is built up as both sides continue to steadily push against each other until there is sudden movement, where both plates quickly slide until they lock up again. This is what causes earthquakes.

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Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

What can you infer about a wave with a short wavelength?

raketka [301]

Answer:

- It can be infer that it has a lower frequency.

<em>In the case of electromagnetic waves.</em>

- A short wavelength means a lower energy,

Explanation:

The wavelength is the distance between two consecutive crests or valleys while the frequency is the number of crests that pass for a specific point in an interval of time.

For example, a person makes laundry once a weak.

In this example, the event is represented by the laundry and the interval of time is once a weak

The velocity of a wave is defined as:

$v = \nu \cdot \lambda$ (1)